Symmetry With a Shifted Center
Even and odd functions are commonly introduced through symmetry about zero: the values at \(t\) and \(-t\) agree or have opposite signs. The same idea works about any real number \(c\). Instead of pairing \(x\) with \(-x\), we pair \(x\) with the point the same distance from \(c\) on the other side: \(2c-x\).
This change of center is useful whenever an integral is taken over an interval centered at \(c\), or when two pieces of an interval are mirror images around \(c\). The essential question is not whether a formula looks even or odd in \(x\), but whether its values agree or cancel at points equally distant from the chosen center.
Equivalently, reflection about \(c\) sends \(x\) to \(2c-x\). Thus even symmetry means \(f(2c-x)=f(x)\), while odd symmetry means \(f(2c-x)=-f(x)\). Taking \(c=0\) recovers the usual definitions.
Translate the Center to Zero
A convenient way to use these definitions is to introduce a centered variable \(t=x-c\). Define \(g(t)=f(c+t)\). Then \(f\) is even about \(c\) precisely when \(g\) is even, and it is odd about \(c\) precisely when \(g\) is odd. The point \(x=c\) becomes \(t=0\), and the reflection pair \(c+t,c-t\) becomes \(t,-t\).
For an interval \([c-r,c+r]\), where \(r>0\), this translation changes the integration limits to \([-r,r]\). The ordinary change-of-variable rule gives \(dx=dt\), so
All symmetry statements below assume the function is integrable on the intervals in question. Continuity on the relevant closed interval is a sufficient condition. The integral rules used here are the familiar ones for even and odd functions on symmetric intervals, applied after translating the center.
Integrals Over an Interval Centered at \(c\)
If \(f\) is even about \(c\), each value to the left of \(c\) matches the value the same distance to the right. The two halves therefore contribute equally. If \(f\) is odd about \(c\), each such pair has sum zero, so the contributions cancel.
Here is a derivation that shows exactly where the symmetry is used. Split the integral at \(c\). In the left-hand integral, set \(t=c-x\). When \(x\) runs from \(c-r\) to \(c\), \(t\) runs from \(r\) down to \(0\), and \(dx=-dt\). Therefore
If \(f\) is even about \(c\), then \(f(c-t)=f(c+t)\). Adding the right-hand half gives
The substitution \(x=c+t\) identifies the last integral with \(\int_c^{c+r}f(x)\,dx\). If \(f\) is odd about \(c\), instead \(f(c-t)=-f(c+t)\). The two halves then sum to zero. This proves both conclusions.
Pairing Mirrored Subintervals
The same reasoning applies to pieces that do not extend all the way from the center to the edge of the interval. Suppose \(0\leq a\leq b\leq r\). The left subinterval \([c-b,c-a]\) and right subinterval \([c+a,c+b]\) are mirror images about \(c\). For an even function, their integrals agree; for an odd function, they are negatives of each other.
For example, the change of variable \(x=2c-u\) maps the left interval from \(x=c-b\) to \(x=c-a\) onto the right interval from \(u=c+b\) down to \(u=c+a\). The negative differential reverses those limits, giving the right-hand integral with the sign determined by whether \(f(2c-u)\) equals \(f(u)\) or \(-f(u)\).
Worked Examples
Worked Example: An Even Function About a Nonzero Center
Evaluate \(\int_1^5 ((x-3)^2+2)\,dx\) by using symmetry about \(c=3\).
The interval \([1,5]\) is centered at \(3\), with \(r=2\). The function depends on \(x-3\) through its square and a constant, so replacing \(x-3\) by its negative does not change its value. It is even about \(3\). The even symmetry formula gives
For the right half, set \(t=x-3\). The limits become \(0\) and \(2\):
Check by evaluating directly on the full interval. An antiderivative is \(\frac{(x-3)^3}{3}+2x\), so
The direct evaluation agrees with the symmetry calculation.
Worked Example: An Odd Function About a Nonzero Center
Evaluate \(\int_{-5}^{1} ((x+2)^3-4(x+2))\,dx\).
The interval has center \(c=-2\) and radius \(r=3\). Let \(t=x+2=x-c\). The integrand becomes \(t^3-4t\), which changes sign when \(t\) is replaced by \(-t\). Thus the function is odd about \(-2\), and the centered-interval rule gives
Check by direct evaluation. An antiderivative is \(\frac{(x+2)^4}{4}-2(x+2)^2\). At the upper endpoint \(x=1\), its value is
At the lower endpoint \(x=-5\), its value is
The definite integral is the upper value minus the lower value, \(\frac{9}{4}-\frac{9}{4}=0\), as predicted.
Worked Example: Equal Integrals on Mirrored Subintervals
Show that the integrals of \(f(x)=(x-2)^2+1\) over \([0,1]\) and \([3,4]\) agree, and find their common value.
The center is \(c=2\). For each \(x\), the reflected point is \(4-x\), and
Thus \(f\) is even about \(2\). The intervals \([0,1]\) and \([3,4]\) are mirrored about \(2\), so their integrals are equal. Evaluate the left one using \(t=x-2\), whose limits are \(-2\) and \(-1\):
For a direct check on the right interval, the same substitution gives limits \(1\) and \(2\):
Both evaluations give \(\frac{10}{3}\), confirming the mirrored-interval rule.
Common Pitfall: The Interval Must Also Be Centered
Odd symmetry about \(c\) guarantees a zero integral over an interval centered at \(c\), not over every interval. A common error is to see a translated odd function and apply cancellation even when the integration limits are not equally spaced around its center.
For instance, \(f(x)=x-2\) is odd about \(2\), but \([0,3]\) is not centered at \(2\). The incorrect calculation would be \(\int_0^3(x-2)\,dx=0\) on the grounds that the function is odd. The actual calculation is
The correction is to check both ingredients: the function must have the stated symmetry about \(c\), and the interval for the zero-integral rule must have the form \([c-r,c+r]\). Here the symmetry center is \(2\), but the endpoints \(0\) and \(3\) are not equally distant from it.
How to Use General-Center Symmetry
When a definite integral appears to have symmetry, identify the center before simplifying. A translated expression such as \(f(x-c)\) often signals that the relevant center is \(c\). Then compare values at \(c+t\) and \(c-t\), and check that the integration interval or the subintervals really are mirror images.
Look for a shifted input such as \(x-c\), or determine the midpoint of the interval endpoints.
Compare \(f(c+t)\) with \(f(c-t)\). Equality gives even symmetry; opposite values give odd symmetry.
Use the full-interval rule only when the limits are \(c-r\) and \(c+r\). For partial pieces, verify that the subintervals are mirrored.
Even symmetry makes paired contributions equal; odd symmetry makes them cancel.
These rules are valuable because they can reduce the amount of integration required and make cancellation visible before evaluating an antiderivative. They also provide a check on a calculation: if two mirrored pieces of an even function produce different integrals, either the symmetry or the arithmetic has been mishandled.
Check Your Understanding
For each question, identify the center and state which symmetry relation applies before calculating.
- Is \(f(x)=(x-4)^4+1\) even or odd about \(4\)?
- If \(f\) is odd about \(c=1\), what is \(\int_{-2}^{4}f(x)\,dx\), assuming the integral exists?
- If \(f\) is even about \(c=3\), how are \(\int_0^1f(x)\,dx\) and \(\int_5^6f(x)\,dx\) related?
- Why does odd symmetry about \(c=2\) alone not imply \(\int_0^3f(x)\,dx=0\)?
- Rewrite \(\int_{c-r}^{c+r}f(x)\,dx\) as an integral over \([-r,r]\) using a centered variable.
